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Physics LibreTexts

15.5: Damped Oscillations

  • Page ID
    4066
  • Learning Objectives

    • Describe the motion of damped harmonic motion
    • Write the equations of motion for damped harmonic oscillations
    • Describe the motion of driven, or forced, damped harmonic motion
    • Write the equations of motion for forced, damped harmonic motion

    In the real world, oscillations seldom follow true SHM. Friction of some sort usually acts to dampen the motion so it dies away, or needs more force to continue. In this section, we examine some examples of damped harmonic motion and see how to modify the equations of motion to describe this more general case.

    A guitar string stops oscillating a few seconds after being plucked. To keep swinging on a playground swing, you must keep pushing (Figure \(\PageIndex{1}\)). Although we can often make friction and other non-conservative forces small or negligible, completely undamped motion is rare. In fact, we may even want to damp oscillations, such as with car shock absorbers.

    A photo of a person on a swing
    Figure \(\PageIndex{1}\): To counteract dampening forces, you need to keep pumping a swing. (credit: Bob Mical)

    Figure \(\PageIndex{2}\) shows a mass m attached to a spring with a force constant k. The mass is raised to a position A0, the initial amplitude, and then released. The mass oscillates around the equilibrium position in a fluid with viscosity but the amplitude decreases for each oscillation. For a system that has a small amount of damping, the period and frequency are constant and are nearly the same as for SHM, but the amplitude gradually decreases as shown. This occurs because the non-conservative damping force removes energy from the system, usually in the form of thermal energy.

    A mass m is suspended from a vertical spring and immersed in a fluid that has viscosity eta. A graph of the damped oscillation shows the displacement x in meters on the vertical axis as a function of time in seconds on the horizontal axis. The range of x is from minus A sub zero to plus A sub zero. The time scale is from zero to 7 T, with tics at increments of T. The displacement is plus A sub zero at time zero and oscillates between positive maxima and negative minima, with each full cycle taking the same time T but the amplitude of the oscillations decreasing with time.
    Figure \(\PageIndex{2}\): For a mass on a spring oscillating in a viscous fluid, the period remains constant, but the amplitudes of the oscillations decrease due to the damping caused by the fluid.

    Consider the forces acting on the mass. Note that the only contribution of the weight is to change the equilibrium position, as discussed earlier in the chapter. Therefore, the net force is equal to the force of the spring and the damping force (\(F_D\)). If the magnitude of the velocity is small, meaning the mass oscillates slowly, the damping force is proportional to the velocity and acts against the direction of motion (\(F_D = −b\)). The net force on the mass is therefore

    $$ma = -bv - kx \ldotp$$

    Writing this as a differential equation in x, we obtain

    $$m \frac{d^{2} x}{dt^{2}} + b \frac{dx}{dt} + kx = 0 \ldotp \label{15.23}$$

    To determine the solution to this equation, consider the plot of position versus time shown in Figure \(\PageIndex{3}\). The curve resembles a cosine curve oscillating in the envelope of an exponential function \(A_0e^{−\alpha t}\) where \(\alpha = \frac{b}{2m}\). The solution is

    $$x(t) = A_{0} e^{- \frac{b}{2m} t} \cos (\omega t + \phi) \ldotp \label{15.24}$$

    It is left as an exercise to prove that this is, in fact, the solution. To prove that it is the right solution, take the first and second derivatives with respect to time and substitute them into Equation 15.23. It is found that Equation 15.24 is the solution if

    $$\omega = \sqrt{\frac{k}{m} - \left(\dfrac{b}{2m}\right)^{2}} \ldotp$$

    Recall that the angular frequency of a mass undergoing SHM is equal to the square root of the force constant divided by the mass. This is often referred to as the natural angular frequency, which is represented as

    $$\omega_{0} = \sqrt{\frac{k}{m}} \ldotp \label{15.25}$$

    The angular frequency for damped harmonic motion becomes

    $$\omega = \sqrt{\omega_{0}^{2} - \left(\dfrac{b}{2m}\right)^{2}} \ldotp \label{15.26}$$

    The figure shows a graph of displacement, x in meters, along the vertical axis, versus time in seconds along the horizontal axis. The displacement ranges from minus A sub zero to plus A sub zero and the time ranges from 0 to 10 T. The displacement, shown by a blue curve, oscillates between positive maxima and negative minima, forming a wave whose amplitude is decreasing gradually as we move far from t=0. The time, T, between adjacent crests remains the same throughout. The envelope, the smooth curve that connects the crests and another smooth curve that connects the troughs of the oscillations, is shown as a pair of dashed red lines. The upper curve connecting the crests is labeled as plus A sub zero times e to the quantity minus b t over 2 m. The lower curve connecting the troughs is labeled as minus A sub zero times e to the quantity minus b t over 2 m.
    Figure \(\PageIndex{3}\): Position versus time for the mass oscillating on a spring in a viscous fluid. Notice that the curve appears to be a cosine function inside an exponential envelope.

    Recall that when we began this description of damped harmonic motion, we stated that the damping must be small. Two questions come to mind. Why must the damping be small? And how small is small? If you gradually increase the amount of damping in a system, the period and frequency begin to be affected, because damping opposes and hence slows the back and forth motion. (The net force is smaller in both directions.) If there is very large damping, the system does not even oscillate—it slowly moves toward equilibrium. The angular frequency is equal to

    $$\omega = \sqrt{\frac{k}{m} - \left(\dfrac{b}{2m}\right)^{2}} \ldotp$$

    As b increases, \(\frac{k}{m} - \left(\dfrac{b}{2m}\right)^{2}\) becomes smaller and eventually reaches zero when b = \(\sqrt{4mk}\). If b becomes any larger, \(\frac{k}{m} - \left(\dfrac{b}{2m}\right)^{2}\) becomes a negative number and \(\sqrt{\frac{k}{m} - \left(\dfrac{b}{2m}\right)^{2}}\) is a complex number.

    The position, x in meters on the vertical axis, versus time in seconds on the horizontal axis, with varying degrees of damping. No scale is given for either axis. All three curves start at the same positive position at time zero. Blue curve a, labeled with b squared is less than 4 m k, undergoes a little over two and a quarter oscillations of decreasing amplitude and constant period. Red curve b, labeled with b squared is equal to 4 m k, decreases at t=0 less rapidly than the blue curve, but does not oscillate. The red curve approaches x=0 asymptotically, and is nearly zero within one oscillation of the blue curve. Green curve c, labeled with b squared is greater than 4 m k, decreases at t=0 less rapidly than the red curve, and does not oscillate. The green curve approaches x=0 asymptotically, but is still noticeably above zero at the end of the graph, after more than two oscillations of the blue curve.
    Figure \(\PageIndex{4}\): The position versus time for three systems consisting of a mass and a spring in a viscous fluid. (a) If the damping is small (b < \(\sqrt{4mk}\)), the mass oscillates, slowly losing amplitude as the energy is dissipated by the non-conservative force(s). The limiting case is (b) where the damping is (b = \(\sqrt{4mk}\)). (c) If the damping is very large (b > \(\sqrt{4mk}\)), the mass does not oscillate when displaced, but attempts to return to the equilibrium position.

    Figure \(\PageIndex{4}\) shows the displacement of a harmonic oscillator for different amounts of damping.

    1. When the damping constant is small, b < \(\sqrt{4mk}\), the system oscillates while the amplitude of the motion decays exponentially. This system is said to be underdamped, as in curve (a). Many systems are underdamped, and oscillate while the amplitude decreases exponentially, such as the mass oscillating on a spring. The damping may be quite small, but eventually the mass comes to rest.
    2. If the damping constant is \(b = \sqrt{4mk}\), the system is said to be critically damped, as in curve (\(b\)). An example of a critically damped system is the shock absorbers in a car. It is advantageous to have the oscillations decay as fast as possible. Here, the system does not oscillate, but asymptotically approaches the equilibrium condition as quickly as possible.
    3. Curve (c) in Figure \(\PageIndex{4}\) represents an overdamped system where \(b > \sqrt{4mk}\). An overdamped system will approach equilibrium over a longer period of time.

    Critical damping is often desired, because such a system returns to equilibrium rapidly and remains at equilibrium as well. In addition, a constant force applied to a critically damped system moves the system to a new equilibrium position in the shortest time possible without overshooting or oscillating about the new position.

    Exercise \(\PageIndex{1}\)

    Why are completely undamped harmonic oscillators so rare?

    Contributors

    • Samuel J. Ling (Truman State University), Jeff Sanny (Loyola Marymount University), and Bill Moebs with many contributing authors. This work is licensed by OpenStax University Physics under a Creative Commons Attribution License (by 4.0).